---
title: "Design main transmission pipes PQ and QR"
subject: "Water Supply Engineering"
chapter: 7
key: "semester_5/water_supply_engineering:pd-c70o"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pd-c70o
site: BCT Engineering
---

# Design main transmission pipes PQ and QR

Water Supply Engineering · Chapter 7: Reservoirs and Distribution System

- Asked in: 2070 Chaitra (old course)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pd-c70o
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-7

## Question

Design suitable diameter of continuous main transmission pipe lines PQ and QR shown below. The average water requirement is 60 lpcd. The storage tank is fixed at point P of R.L. 325.0 m. The water is distributed only after the point Q for a population of 2500 and R for a population of 21000. If the minimum pressure head of water is to be 15.0 m at Q and R. Assume Hazen Williams' coefficient C as 100 and other necessary requirements if necessary.

[Figure: P (RL 325 m) to Q (RL 185 m), length 600 m; Q to R (RL 45 m), length 250 m.]

## Short answer

Use Hazen-Williams equation. For PQ, calculated 123.6 mm, adopt 150 mm, velocity 2.77 m/s. For QR, calculated 88.4 mm, adopt 150 mm, velocity 2.48 m/s. Recommend break pressure tank at Q.

## Answer

Average demand 60 lpcd. The peak factor is not stated; 3 is assumed. C = 100 and residual head = 15 m at Q and R. HGL at P = 325 m. Pipe PQ carries the demand of Q and R: (2500 + 21000) = 23,500 people, and pipe QR carries the 21,000 people at R.

- $Q_{PQ} = 23500 \times 60 \times 3/86400 = 48.96$ lps
- $Q_{QR} = 21000 \times 60 \times 3/86400 = 43.75$ lps

#### Method
Use the Hazen-Williams equation in SI units:
$$
h_f = \frac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}
$$
with $Q$ in m³/s, $L$ and $D$ in m. For each pipe the available head is (HGL at upstream end − ground RL of the tap − required residual head). The gradient $S = H/L$ gives the calculated diameter, the next larger commercial size is adopted, and the real head loss and residual head are then checked. Commercial sizes (nominal bore): 15, 20, 25, 32, 40, 50, 65, 80, 100, 125, 150, 200 mm and so on. Constant C = 100.

#### Pipe PQ
- Discharge $Q = 48.958$ lps, length $L = 600$ m
- HGL at P = 325.00 m; ground RL at Q = 185 m; residual head required = 15 m
- Available head $H = 325.00 - 185 - 15 = 125.00$ m
- Gradient $S = H/L = 125.00/600 = 0.20833$
- From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 123.6$ mm, adopt commercial size **150 mm** (raised from the head-based size because velocity would exceed 3 m/s)
- Actual $h_f = 48.788$ m, HGL at Q = 276.21 m, residual head = 91.21 m (≥ 15 m, OK)
- Velocity $v = Q/A = 2.77$ m/s

#### Pipe QR
- Discharge $Q = 43.750$ lps, length $L = 250$ m
- HGL at Q = 276.21 m; ground RL at R = 45 m; residual head required = 15 m
- Available head $H = 276.21 - 45 - 15 = 216.21$ m
- Gradient $S = H/L = 216.21/250 = 0.86485$
- From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 88.4$ mm, adopt commercial size **150 mm** (raised from the head-based size because velocity would exceed 3 m/s)
- Actual $h_f = 16.506$ m, HGL at R = 259.71 m, residual head = 214.71 m (≥ 15 m, OK)
- Velocity $v = Q/A = 2.48$ m/s

#### Summary
| Pipe | Q (lps) | L (m) | Calc. D (mm) | Adopted D (mm) | h_f (m) | Residual (m) | v (m/s) |
|---|---|---|---|---|---|---|---|
| PQ | 48.9583 | 600 | 123.6 | 150 | 48.79 | 91.21 | 2.77 |
| QR | 43.75 | 250 | 88.4 | 150 | 16.51 | 214.71 | 2.48 |

The static head at Q is 325 − 185 = 140 m and at R is 325 − 45 = **280 m**, which is too high for ordinary pipes (and the residual heads of 91 m and 215 m are far more than needed). A **break pressure tank at Q** (RL 185 m) should be provided: the pressure is reduced to zero there, QR then has a static head of 140 m only, and the head available for QR becomes about 185 − 45 − 15 = 125 m, but the velocity limit of 3 m/s (125 mm gives 3.6 m/s) still requires 150 mm.

**Answer: PQ = 150 mm and QR = 150 mm (with a BPT recommended at Q).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which population does pipe PQ serve?
   A. Only Q (2500)
   B. Only R (21000)
   C. Both Q and R (23500)
   D. Neither; it only stores water
   Answer: C. PQ carries the demand of Q and R together, 23,500 people, while QR carries only R. (Key point: Discharge in PQ)

2. The head-based sizes were 123.6 mm (PQ) and 88.4 mm (QR), yet 150 mm was adopted. Why?
   A. Velocity would exceed 3 m/s with the smaller sizes
   B. Residual head was too low
   C. Pipes below 150 mm do not exist
   D. C was changed
   Answer: A. Head is plentiful, so the smaller sizes would give velocity above 3 m/s; 150 mm keeps it below. (Key point: Why 150 mm)

3. Why is a break pressure tank recommended at Q?
   A. To increase flow in QR
   B. To raise pressure at R
   C. To store water for fire
   D. Static heads (140 m at Q, 280 m at R) are too high for ordinary pipes; the tank resets pressure to zero
   Answer: D. A BPT at Q reduces pressure to zero so QR only sees 140 m static head. (Key point: Break pressure tank)

4. The peak factor was not given. What was done?
   A. Taken as 1
   B. A value of 3 was assumed
   C. Question left unsolved
   D. Taken as 10
   Answer: B. The answer assumes a peak factor of 3, giving 48.96 and 43.75 lps. (Key point: Assumed peak factor)

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